Ass 2: Template
Shone George Kutty Renjan
Monday, October 07, 2024
Questions
Q 1
# Question 1 Table
table = matrix(c(0.9212, 0.7455, 0.9932, 0.4482, 1.0, 0.7703), nrow=3, ncol=2, byrow=TRUE, dimnames = list(c('Match', 'Similar Distracter', 'Non-similar Distracter'), c('Fingerprint experts', 'Novices')))
table## Fingerprint experts Novices
## Match 0.9212 0.7455
## Similar Distracter 0.9932 0.4482
## Non-similar Distracter 1.0000 0.7703a.
# Given a pair of matched prints. What is the probability that an expert will fail to identify the match?
failmatchprob = (1-table[1,1])/(sum(table["Match",]))
failmatchprob## [1] 0.04727905b.
## Given a pair of matched prints, what is the probability that a novice will fail to identify the match?
novicefailmatchprob = (1- table[1,2])/(sum(table["Match",]))
novicefailmatchprob## [1] 0.1526969c.
Assume the study included 10 participants, 5 experts and 5 novices. Suppose that a pair of matched prints are presented to a randomly selected study participant and the participant fails to identify the match. Is the participant more likely to be an expert or a novice?
Novice, because the likelihood of a novice failing to find the match is higher than that of an expert
Q 2
table1 = matrix(c(50, 9, 50, 891), ncol=2, byrow=TRUE, dimnames = list(c('Positive', 'Negative'), c('Users', 'Non-users')))
addmargins(table1)## Users Non-users Sum
## Positive 50 9 59
## Negative 50 891 941
## Sum 100 900 1000a.
# Given the athlete is a user, find the probability that a drug test for testosterone will yield a positive result. This probability represents the sensitivity of the drug test.
posuser = (table1[1,1])/(sum(table1[,"Users"]))
posuser## [1] 0.5b.
#Given the athlete is a nonuser, find the probability that a drug test for testosterone will yield a negative result. This probability represents the specificity of the drug test.)
negnonuser = table1[2,2]/(sum(table1[,"Non-users"]))
negnonuser## [1] 0.99c.
# IF an athlete tests positive for testosterone, use Bayes' Rule to find the probability that the athlete is really doping. (This probability represents the positive predictive value of the drug test.)
## Bayes' Rule = (P(Positive | Doping)*P(Doping))/P(Positive)
totalNonuserUser = sum(table1[, "Users"]) + sum(table1[, "Non-users"])
totalpositive=sum(table1["Positive",])
priorProbDope = sum(table1[,"Users"]) / totalNonuserUser
totalProbpostive = totalpositive/totalNonuserUser
reallyDoping = (posuser*priorProbDope)/totalProbpostive
reallyDoping## [1] 0.8474576Q 3
Base Case: For a single set with n₁ elements, we can create n₁ different samples by choosing one element. Thus, for k=1, the principle holds. Inductive Step: Assume the principle holds for k sets, i.e., the number of samples is n₁ * n₂ * … * nₖ. For k+1 sets, consider the samples from k sets (n₁ * n₂ * … * nₖ) and pair each with an element from the (k+1)th set (nₖ₊₁). This results in (n₁ * n₂ * … * nₖ) * nₖ₊₁ samples. Therefore, by induction, the Multiplicative Principle holds for all k ∈ ℕ. for all𝑘∈ℕk∈N
Q 4
knitr::include_graphics("/Users/shonerenjan/Desktop/ASM/Assignments/Assignment2/4.jpg")
Q 5
knitr::include_graphics("/Users/shonerenjan/Desktop/ASM/Assignments/Assignment2/5.png")
Q 6
knitr::include_graphics("/Users/shonerenjan/Desktop/ASM/Assignments/Assignment2/6.jpg")
Q 7
table2 = matrix(c(0.09, 0.3, 0.37, 0.2, 0.04), nrow = 1, byrow = TRUE, dimnames = list(c('p(y)'), c('0', '1', '2', '3', '4')))
table2## 0 1 2 3 4
## p(y) 0.09 0.3 0.37 0.2 0.04a.
###.Verify that the probabilities for Y in the table sum to 1.
sum(table2['p(y)',])## [1] 1b.
### Find the probability that three or four of the homes in the sample have a dust mite level that exceeds 2 g/g.
probhouse3orprobhouse4 = table2[1,4]+ table2[1,5]
probhouse3orprobhouse4## [1] 0.24c.
### Find the probability that fewer than two homes in the sample have a dust mite level that exceeds 2 g/g.
probhouse0orprobhouse1 = table2[1,1]+ table2[1,2]
probhouse0orprobhouse1## [1] 0.39Q 8
# Creating a matrix with 21 rows and 2 columns
number_of_apps <- c(0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20)
probabilities <- c(0.17, 0.10, 0.11, 0.11, 0.10, 0.10, 0.07, 0.05, 0.03, 0.02, 0.02, 0.02, 0.02, 0.02, 0.01, 0.01, 0.01, 0.01, 0.01, 0.005, 0.005)
table4 <- matrix(c(number_of_apps, probabilities), ncol = 2, byrow = FALSE, dimnames = list(NULL, c('Number of apps used', 'p(y)')))
table4## Number of apps used p(y)
## [1,] 0 0.170
## [2,] 1 0.100
## [3,] 2 0.110
## [4,] 3 0.110
## [5,] 4 0.100
## [6,] 5 0.100
## [7,] 6 0.070
## [8,] 7 0.050
## [9,] 8 0.030
## [10,] 9 0.020
## [11,] 10 0.020
## [12,] 11 0.020
## [13,] 12 0.020
## [14,] 13 0.020
## [15,] 14 0.010
## [16,] 15 0.010
## [17,] 16 0.010
## [18,] 17 0.010
## [19,] 18 0.010
## [20,] 19 0.005
## [21,] 20 0.005a.
## Show that the properties of a probability distribution for a discrete random variable are satisfied.
non_negativity_check <- all(probabilities >= 0)
non_negativity_check ## [1] TRUEnormalization_check <- sum(probabilities)
normalization_check ## True## [1] 1b.
# Find P(Y ≥ 10).
P_Y10 <- sum(probabilities[11:length(probabilities)])
P_Y10## [1] 0.14c.
## Find the mean and variance of Y.
mean(probabilities)## [1] 0.04761905(sd(probabilities))^2## [1] 0.002316548d.
## Give an interval that will contain the value of Y with a probability of at least .75.
sum(table4[,"p(y)"])## [1] 1Q 9
\[n = 25\] \[p = .70\] ### a.
# Find P(Y = 10).
dbinom(10,25,0.7)## [1] 0.001324897b.
# Find P(Y ≤ 5).
pbinom(5,25,0.7)## [1] 3.457444e-07c.
# Find the mean u and standard deviation o for Y.
μ= 25*.7
μ## [1] 17.5σ= sqrt(25 *0.7 *0.3)
σ## [1] 2.291288d.
Out of 25 students who receive their PhDs, around 18 are non-foreign nationals, with a standard deviation of 2.29.
Q 10
a.
## What is the probability that exactly 5 trains are assigned to each of the 10 tracks?
## 50!/40!
assignedprob=((factorial(50))/(factorial(45)))/((factorial(50))/(factorial(40)))
assignedprob## [1] 6.820767e-09b.
## A track is considered underutilized if fewer than 2 trains are assigned to the track during the day. Find the probability that Track #1 is underutilized.
probtrackunderutilized = dbinom(0,50,0.1)+dbinom(1,50,0.1)
probtrackunderutilized## [1] 0.03378586Q 11
a.
Give a formula for the probability distribution of Y. (P(Y == y) == p * q^(y - 1))
p=(12+6+4+18)/(100)
p## [1] 0.4b.
\[ E(Y)= p/1\] What is E(Y)? Interpret the result. This formula calculates the expected number of trials needed to encounter a consumer who provides a reason for purchasing the product that differs from the information presented on its label or packaging.
c.
# Find P(Y = 1).
dgeom(1,p)## [1] 0.24d.
1-pgeom(2,p)## [1] 0.216Q 12
a.
## In a random sample of 10 of the 209 facilities, what is the expected number in the sample that treats hazardous waste on-site? Interpret this result.
p=8/209
n=10
E=p*n
cat("Treating onsite: ",E)## Treating onsite: 0.3827751There at least 1 facility treats hazardous waste on site.
b.
## Find the probability that 4 of the 10 selected facilities treat hazardous waste on-site.
dhyper(4,8,201,10)## [1] 0.0001688459Q 13
a.
Find the variance of Y. \[ E(Y)=0.03 \] \[ Var(Y)= E(Y) =0.03 \]
b.
Discuss the conditions that would make the researchers’ Poisson assumption plausible.
The experiment involves recording how often the event Y, which is rare, happens within a specific period, area, volume, or any other defined measurement unit. The likelihood of the event occurring is consistent across all units of time, area, or volume, and these units do not overlap. The occurrence of events in one unit is independent of their occurrence in any other unit.
c.
## What is the probability that a deep-draft U.S. flag vessel will have no casualties in a 3-year time period?
dpois(0,0.03)## [1] 0.9704455Q 14
a.
findc <- function(c) {
integrate(function(y) (c * (2 - y)) / 10, lower = 0, upper = 1)$value
}
cvalue <- 2 / 3 # Use 'cvalue' here
integral_value <- findc(cvalue) # Use 'cvalue' here as well
cat("c =", cvalue)## c = 0.6666667b.
knitr::include_graphics("/Users/shonerenjan/Desktop/ASM/Assignments/Assignment2/15.jpg")
c.
F <- function(y){(cvalue)*(2*y-((y^2)/2))}d.
F(0.6)-F(0.1)## [1] 0.55Q 15
a-c
knitr::include_graphics("/Users/shonerenjan/Desktop/ASM/Assignments/Assignment2/16.jpg")
Q 16
\[𝜇=50\] \[𝜎=3.2\]
a.
# exceeding 45 milligrams per liter.
1-pnorm(45,50,3.2)## [1] 0.9409149b.
# b. below 55 milligrams per liter.
pnorm(55,50,3.2)## [1] 0.9409149c.
# between 51 and 52 milligrams per liter.
pnorm(52,50,3.2)-pnorm(51,50,3.2)## [1] 0.1113448Q 17
a.
#Find the probability that the rating will fall between 500 and 700 points.
pnorm(700,605,185)-pnorm(500,605,185)## [1] 0.4110396b.
## Find the probability that the rating will fall between 400 and 500 points.
pnorm(500,605,185)-pnorm(400,605,185)## [1] 0.1512568c.
# Find the probability that the rating will be less than 850 points.
pnorm(850,605,185)## [1] 0.9073023d.
# Find the probability that the rating will exceed 1,000 points.
1-pnorm(1000,605,185)## [1] 0.01637499e.
# What rating will only 10% of the crash-tested cars exceed?
qnorm(0.9,605,185)## [1] 842.087